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ivann1987 [24]
2 years ago
6

Which substances will make a salt when combined? (2 points) Vinegar and antacids Soda and wine Soap and ammonia Fertilizer and d

etergent
Chemistry
2 answers:
SCORPION-xisa [38]2 years ago
7 0

vinegar and antacids.

Ber [7]2 years ago
5 0

Answer:

A) vinegar and antacids

Explanation:

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How many grams of aluminum chloride must decompose to produce 78.3 milliliters of aluminum metal, if the density of aluminum is
Vilka [71]
1mol aluminium chloride gives 1mol aluminium and 3mol chloride
density equals mass divided by volume
d=m/v
m=v*d
=78.3*2.7
=211.41grams
7 0
3 years ago
When an atom loses electrons it becomes___
murzikaleks [220]
<span>It becomes an </span>ion<span>. An atom that gains a negative electron, it becomes a negative </span>ion<span>. If it loses an electron it becomes a positive </span>ion<span>.</span>

Hope this helps!

8 0
2 years ago
What is a scientific question?
Elena L [17]

Answer:

O c. One that is answered through observation

I used process of elminations so

Explanation:

One that everyone agrees one doesnt mean it's accurate.

Only the one that scientists ask doesnt make sense

One that never changes and One that is answered through observation.... So C

5 0
3 years ago
PLEASE ANSWEER THIS HURRY HURRRRRRRRRYY
klemol [59]
The first one would be it
6 0
3 years ago
Arrange the following H atom electron transitions in order of decreasing wavelength of the photon absorbed or emitted:
Katyanochek1 [597]

The decreasing order of wavelengths of the photons emitted or absorbed by the H atom is : b → c → a → d

Rydberg's formula :

                                   \frac{1}{\lambda} = R_h (\frac{1}{n_1^2}-\frac{1}{n_2^2} ),

where  λ is the wavelength of the photon emitted or absorbed from an H atom electron transition from n_1 to n_2 and R_h = 109677 is the Rydberg Constant. Here n_1 and  n_2 represents the transitions.

(a) n_1 =2 to n_2 = infinity

            \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{\infty^2}  ) = 109677/4     [since 1/infinity = 0] Therefore, \lambda = 4 / 109677 = 0.00003647 m

(b) n_1=4 to  n_2 = 20

           \frac{1}{\lambda} = 109677\times (\frac{1}{4^2}-\frac{1}{20^2}  ) = 6580.62

Therefore,  \lambda = 1 / 6580.62 = 0.000152 m

(c) n_1=3 to  n_2 = 10

          \frac{1}{\lambda} = 109677\times (\frac{1}{3^2}-\frac{1}{10^2}  ) = 11089.56

Therefore,  \lambda = 1 / 11089.56 = 0.00009 m

(d)  n_1=2 to  n_2 = 1

          \frac{1}{\lambda} = 109677\times (\frac{1}{2^2}-\frac{1}{1^2}  ) = - 82257.75

Therefore,  \lambda = 1 /82257.75  = - 0.0000121 m  

[Even though there is a negative sign, the magnitude is only considered because the sign denotes that energy is emitted.]

So the decreasing order of wavelength of the photon absorbed or emitted is b → c → a → d.

Learn more about the Rydberg's formula athttps://brainly.com/question/14649374

#SPJ4

8 0
1 year ago
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